Consider the following statements :

Assertion $(A)$ : The circle ${x^2} + {y^2} = 1$ has exactly two tangents parallel to the $x$ - axis

Reason $(R)$ : $\frac{{dy}}{{dx}} = 0$ on the circle exactly at the point $(0, \pm 1)$.

Of these statements

A tangent to the circle ${x^2} + {y^2} = 5$at the point $(1,-2)$ the circle ${x^2} + {y^2} - 8x + 6y + 20 = 0$

Let a circle $C$ of radius $5$ lie below the $x$-axis. The line $L_{1}=4 x+3 y-2$ passes through the centre $P$ of the circle $C$ and intersects the line $L _{2}: 3 x -4 y -11=0$ at $Q$. The line $L _{2}$ touches $C$ at the point $Q$. Then the distance of $P$ from the line $5 x-12 y+51=0$ is

The tangent$(s)$ from the point of intersection of the lines $2x -3y + 1$ = $0$ and $3x -2y -1$ = $0$ to circle $x^2 + y^2 + 2x -4y$ = $0$ will be -

The equations of the tangents drawn from the point $(0, 1)$ to the circle ${x^2} + {y^2} - 2x + 4y = 0$ are

If the equation of the tangent to the circle ${x^2} + {y^2} - 2x + 6y - 6 = 0$ parallel to $3x - 4y + 7 = 0$ is $3x - 4y + k = 0$, then the values of $k$ are